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TomLikesPhysics
Group Title
Where is the gradient of phi
a) perpendicular
b) parallel
to the zaxis?
 one year ago
 one year ago
TomLikesPhysics Group Title
Where is the gradient of phi a) perpendicular b) parallel to the zaxis?
 one year ago
 one year ago

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TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
I guess I have to set something zero for a) and perhaps in b) z should be constant?
 one year ago

myko Group TitleBest ResponseYou've already chosen the best response.2
use dot product. If need more help tell. :)
 one year ago

TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
I do the dot thing to get the parellel part and a cross product for the perpendicular part?
 one year ago

myko Group TitleBest ResponseYou've already chosen the best response.2
yes. to find where it is perpendicular to z axis solve this equation: nabla f . (0,0,1)=0 for paralel : nabla f . (0,0,1)= nabla f
 one year ago

myko Group TitleBest ResponseYou've already chosen the best response.2
other way would be finding maximum and minimum of nabla f
 one year ago

TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
So for the parallel part I dont use a any vector but the vector (0,0,1) and dot this vector with nabla?
 one year ago

TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
for a) I ended up with z^2=xy Is that my final answert for a set of points for which the gradient is parallel to the zaxis?
 one year ago

TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
for b) I got now y^2xz=x^2yz This looks also not very good as a final answer.
 one year ago

myko Group TitleBest ResponseYou've already chosen the best response.2
for b): 3(x^2yz,y^2xz,z^2xy).(0,0,1)=3sqrt((x^2yz)^2+(y^2xz)^2+(z^2xy)^2)=> (z^2xy)=sqrt((x^2yz)^2+(y^2xz)^2+(z^2xy)^2) now should try to simplify this somehow
 one year ago

myko Group TitleBest ResponseYou've already chosen the best response.2
noticing that it will happen when : (x^2yz)^2=0 and (y^2xz)^2=0
 one year ago

myko Group TitleBest ResponseYou've already chosen the best response.2
so x^2=yz and y^2=xz from 1º: y=x^2/z putting this in 2º: x^4/z^2=xz => x^3=z^3 =>x=z=y which is a line passing through the origin forming 45º with every axis
 one year ago

myko Group TitleBest ResponseYou've already chosen the best response.2
@TomLikesPhysics
 one year ago

TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
So x, y, and z can be any number but as long as the have all the same value the gradient of phi is parallel to the zaxis?
 one year ago

TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
:( c) When is the gradient zero. There I already calculated that the gradient would be zero for x=y=z so it seems kind of weird that this is the same solution for the gradient to be parallel to the zaxis.
 one year ago

TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
But you calculated x=y=z so how can x=y=0 and z= anything?
 one year ago

myko Group TitleBest ResponseYou've already chosen the best response.2
Ya that's true. Forget it.
 one year ago

myko Group TitleBest ResponseYou've already chosen the best response.2
On the other hand as much as I remmeber vector that is 0 is also considered perpendicular or parallel to anything
 one year ago

TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
This is confusing. Did I at least got a) right? I ended up with z^2=xy but that looks kind of weird for a final answert, but I don´t know if I can simplify or clarify that anymore.
 one year ago

myko Group TitleBest ResponseYou've already chosen the best response.2
Yes it's right. Here you got the graph of that: http://www.wolframalpha.com/input/?i=z%5E2%3Dxy+
 one year ago

myko Group TitleBest ResponseYou've already chosen the best response.2
I think part b) is ok too. Not sure tough, how to explain that it is same as for c)
 one year ago

TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
Hmmmm ... I am still pretty confused but if you think this is alright than I will stick to it. Thank you for your help, especially since this took so long. Thanks a lot, myko.
 one year ago

myko Group TitleBest ResponseYou've already chosen the best response.2
Glad to help. Sry for not clarifying it till the end...
 one year ago
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